Step-by-Step Answers: Acids, Bases & Salts NCERT Solutions Class 10

Understanding pH and Litmus Behaviour

Conceptual Theory:

The pH scale is used to measure how acidic or basic a solution is. It ranges from 0 to 14:

pH < 7 → Acidic solution
pH = 7 → Neutral solution (e.g., pure water)
pH > 7 → Basic (alkaline) solution

Litmus is a natural indicator:

Acids turn blue litmus → red
Bases turn red litmus → blue

Solution Roadmap:

Step 1: Identify the nature of solution using litmus behaviour
Step 2: Relate nature (acid/base) with pH scale
Step 3: Match correct pH value from given options

1
4
5
10

Step-by-Step Solution:
Step 1: The solution turns red litmus blue.
⇒ This indicates the solution is basic (alkaline).

Step 2: For basic solutions, pH value is always:
\[ \text{pH} > 7 \]
Step 3: Now evaluate each option:

1 → \(\text{pH} = 1\), strongly acidic ❌
4 → \(\text{pH} = 4\), weakly acidic ❌
5 → \(\text{pH} = 5\), weakly acidic ❌
10 → \(\text{pH} = 10\), basic ✔

Step 4: Only one value satisfies the condition of being basic.

Final Answer: 10

Exam Significance:

This question tests the direct correlation between pH scale and indicator behaviour, a frequently asked concept in CBSE board exams.
Competitive exams (NTSE, Olympiads, SSC, Railways) often include similar MCQs requiring quick identification of acidic/basic nature.
Understanding this helps in solving higher-level numericals involving pH, hydrogen ion concentration, and neutralisation reactions.

Acids Reacting with Carbonates

Conceptual Theory:

Egg-shells are primarily composed of calcium carbonate (CaCO₃). Carbonates react with acids to produce:

Salt
Water
Carbon dioxide gas (CO₂)

The CO₂ gas produced has a characteristic test:

CO₂ turns lime-water milky due to formation of CaCO₃ precipitate

Solution Roadmap:

Step 1: Identify substance in egg-shell (CaCO₃)
Step 2: Recall reaction of carbonates with acids
Step 3: Identify gas evolved (CO₂) and its test
Step 4: Select the substance that is an acid from given options

NaCl
HCl
LiCl
KCl

Step-by-Step Solution:
Step 1: Egg-shell contains calcium carbonate:
\[ \mathrm{CaCO_3} \]
Step 2: Carbonates react only with acids to produce CO₂ gas:
\[ \mathrm{Carbonate + Acid \rightarrow Salt + H_2O + CO_2} \]
Step 3: The gas produced turns lime-water milky.
⇒ This confirms the gas is CO₂.

Step 4: Therefore, the solution must be an acid.

Step 5: Check given options:

NaCl → Neutral salt ❌
HCl → Acid ✔
LiCl → Neutral salt ❌
KCl → Neutral salt ❌

Step 6: Only HCl satisfies the condition.

Final Answer: HCl

Chemical Reaction (Balanced):
\[ \mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2} \]

Exam Significance:

This is a standard reaction used to identify carbonates and acids in qualitative analysis.
Frequently asked in CBSE board exams in both MCQ and short-answer formats.
Important for competitive exams where gas identification and reaction prediction are tested.
Builds foundation for chapters involving chemical reactions and gas tests.

Neutralisation and Volume Relationships

Conceptual Theory:

Neutralisation is the reaction between an acid and a base to form salt and water.

\[ \mathrm{Acid + Base \rightarrow Salt + Water} \]

When two solutions react completely, the number of moles of acid equals the number of moles of base. For solutions of same concentration, we can use:

\[ n_1 V_1 = n_2 V_2 \]

Since both solutions remain the same, concentration is constant, hence:

\[ \frac{V_{\mathrm{HCl}}}{V_{\mathrm{NaOH}}} = \text{constant} \]

Solution Roadmap:

Step 1: Write given data from first condition
Step 2: Use proportionality between volumes
Step 3: Substitute new NaOH volume
Step 4: Solve step-by-step for required HCl volume

4 mL
8 mL
12 mL
16 mL

Step-by-Step Solution:
Step 1: Given initial condition:
Volume of NaOH = 10 mL
Volume of HCl = 8 mL

Step 2: Since concentration is same, ratio remains constant:
\[ \frac{V_{\mathrm{HCl}}}{V_{\mathrm{NaOH}}} = \frac{8}{10} \]
Step 3: For new condition:
Volume of NaOH = 20 mL
Let required HCl = \(V\)

Step 4: Apply proportionality:
\[ \frac{V}{20} = \frac{8}{10} \]
Step 5: Solve step-by-step:
\[ V = \frac{8 \times 20}{10} \]
\[ V = \frac{160}{10} \]
\[ V = 16 \text{ mL} \]

Step 6: Final conclusion:
Required volume of HCl = 16 mL

Balanced Chemical Reaction:
\[ \mathrm{NaOH + HCl \rightarrow NaCl + H_2O} \]

The molar ratio between NaOH and HCl is 1:1, hence doubling NaOH volume requires doubling HCl volume.

Exam Significance:

Very common CBSE board question testing numerical application of neutralisation.
Important for competitive exams involving mole concept and titration logic.
Helps in understanding real laboratory techniques like acid-base titration.
Strengthens proportional reasoning, which is frequently tested in MCQs.

Role of Antacids in Neutralising Stomach Acid

Conceptual Theory:

The stomach secretes hydrochloric acid (HCl) to aid digestion. However, excess acid can cause indigestion (dyspepsia), leading to symptoms like burning sensation, pain, and discomfort.

To control excess acidity, substances that are basic in nature are used. These are called antacids.

Antacids neutralise excess acid using a neutralisation reaction:

\[ \mathrm{Acid + Base \rightarrow Salt + Water} \]

Solution Roadmap:

Step 1: Identify cause of indigestion (excess acid)
Step 2: Determine type of substance needed (base)
Step 3: Match with given medicine categories

Antibiotic
Analgesic
Antacid
Antiseptic

Step-by-Step Solution:
Step 1: Indigestion is caused due to excess acid (HCl) in the stomach.

Step 2: To reduce acidity, a base is required.

Step 3: A substance that neutralises acid is called an antacid.

Step 4: Analyse options:

Antibiotic → Kills bacteria ❌
Analgesic → Relieves pain ❌
Antacid → Neutralises acid ✔
Antiseptic → Prevents infection ❌

Step 5: Only one option directly neutralises excess stomach acid.

Final Answer: Antacid

Example Neutralisation Reaction:
\[ \mathrm{Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O} \]

Exam Significance:

Direct conceptual question frequently asked in CBSE board MCQs.
Important for linking chemistry with everyday life applications.
Common in competitive exams testing application of neutralisation reactions.
Helps in understanding real-life use of bases in medicine.

Acids Reacting with Metals (Salt + Hydrogen Formation)

Conceptual Theory:

Metals react with dilute acids to produce a salt and hydrogen gas. This is a displacement reaction where hydrogen is displaced from the acid.

\[ \mathrm{Metal + Dilute\ Acid \rightarrow Salt + H_2 \uparrow} \]

The reactivity of the metal determines how fast the reaction occurs. Metals like Zn, Mg, Al, and Fe react readily with dilute acids.

Solution Roadmap:

Step 1: Identify metal and acid involved
Step 2: Write word equation (metal + acid → salt + hydrogen)
Step 3: Write skeletal chemical equation
Step 4: Balance atoms step-by-step

dilute sulphuric acid reacts with zinc granules.
dilute hydrochloric acid reacts with magnesium ribbon.
dilute sulphuric acid reacts with aluminium powder.
dilute hydrochloric acid reacts with iron filings.

1. Dilute sulphuric acid + Zinc

Step 1: Word equation:
Zinc + Sulphuric acid → Zinc sulphate + Hydrogen

Step 2: Skeletal equation:
\[ \mathrm{Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2} \]
Step 3: Check balance:
Zn = 1, H = 2, SO₄ = 1 on both sides ✔

Final balanced equation:
\[ \mathrm{Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2} \]

2. Dilute hydrochloric acid + Magnesium

Step 1: Word equation:
Magnesium + Hydrochloric acid → Magnesium chloride + Hydrogen

Step 2: Skeletal equation:
\[ \mathrm{Mg + HCl \rightarrow MgCl_2 + H_2} \]
Step 3: Balance chlorine first:
\[ \mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2} \]
Step 4: Check atoms:
Mg = 1, Cl = 2, H = 2 ✔

Final balanced equation:
\[ \mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2} \]

3. Dilute sulphuric acid + Aluminium

Step 1: Word equation:
Aluminium + Sulphuric acid → Aluminium sulphate + Hydrogen

Step 2: Skeletal equation:
\[ \mathrm{Al + H_2SO_4 \rightarrow Al_2(SO_4)_3 + H_2} \]
Step 3: Balance aluminium atoms:
\[ \mathrm{2Al + H_2SO_4 \rightarrow Al_2(SO_4)_3 + H_2} \]
Step 4: Balance sulphate groups:
\[ \mathrm{2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + H_2} \]
Step 5: Balance hydrogen:
Left side H = 6 → Right side needs 3H₂
\[ \mathrm{2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2} \]

Final balanced equation:
\[ \mathrm{2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2} \]

4. Dilute hydrochloric acid + Iron

Step 1: Word equation:
Iron + Hydrochloric acid → Iron(II) chloride + Hydrogen

Step 2: Skeletal equation:
\[ \mathrm{Fe + HCl \rightarrow FeCl_2 + H_2} \]
Step 3: Balance chlorine:
\[ \mathrm{Fe + 2HCl \rightarrow FeCl_2 + H_2} \]
Step 4: Check atoms:
Fe = 1, Cl = 2, H = 2 ✔

Final balanced equation:
\[ \mathrm{Fe + 2HCl \rightarrow FeCl_2 + H_2} \]

Exam Significance:

Very important for CBSE board exams: frequently asked to write word and balanced equations.
Tests understanding of metal reactivity and displacement reactions.
Common in competitive exams requiring balancing of chemical equations.
Helps in identifying hydrogen gas evolution in laboratory experiments.

Why Alcohols and Glucose Are Not Acids (Activity-Based Proof)

Conceptual Theory:

Acids are substances that produce hydrogen ions \(\mathrm{(H^+)}\) in aqueous solution. The presence of free \(\mathrm{H^+}\) ions is responsible for acidic properties such as:

Changing blue litmus to red
Conducting electricity in aqueous solution

Although alcohols and glucose contain hydrogen atoms, they do not ionise in water to release \(\mathrm{H^+}\) ions. Hence, they are not acids.

Solution Roadmap:

Step 1: Take solutions of alcohol, glucose, and an acid
Step 2: Test with litmus paper
Step 3: Compare observations
Step 4: Conclude based on ionisation behavior

Activity (Step-by-Step):

Step 1: Take three clean test tubes and label them A, B, and C.

Step 2: Add solutions:
Test Tube A → Ethanol (alcohol)
Test Tube B → Glucose solution
Test Tube C → Dilute HCl (acid)

Step 3: Test with blue litmus paper:
A → No change
B → No change
C → Turns red

Step 4: Test with red litmus paper:
A → No change
B → No change
C → No change (already red)

Step 5: (Optional) Check conductivity:
Only HCl solution conducts electricity due to presence of ions.

Observations:
Alcohol and glucose do not change litmus colour and do not produce ions.
HCl shows acidic behaviour.

Conclusion:
Acids produce \(\mathrm{H^+}\) ions in aqueous solution:
\[ \mathrm{HCl \rightarrow H^+ + Cl^-} \]
Alcohol and glucose do not ionise:
\[ \mathrm{C_2H_5OH \nrightarrow H^+ + ...} \]
Therefore, even though they contain hydrogen, they are not acids.

Exam Significance:

Very important CBSE board question based on activity-based explanation.
Frequently asked in 2–3 mark questions requiring reason + experiment.
Helps distinguish between presence of hydrogen vs release of H⁺ ions.
Important for competitive exams testing conceptual clarity of acids.

Electrical Conductivity of Water (Role of Ions)

Conceptual Theory:

Electrical conductivity in aqueous solutions occurs due to the presence of free ions. Pure water (distilled water) has extremely low ion concentration due to very slight self-ionisation:

\[ \mathrm{H_2O \rightleftharpoons H^+ + OH^-} \]

The concentration of ions is very small \(\left(10^{-7}\,\mathrm{M}\right)\), so distilled water practically does not conduct electricity.

Rainwater, however, dissolves gases like carbon dioxide from the atmosphere:

\[ \mathrm{CO_2 + H_2O \rightarrow H_2CO_3} \]

This carbonic acid further ionises to produce ions:

\[ \mathrm{H_2CO_3 \rightarrow H^+ + HCO_3^-} \]

Solution Roadmap:

Step 1: Recall that conduction requires ions
Step 2: Analyse ion concentration in distilled water
Step 3: Analyse composition of rainwater
Step 4: Compare both cases

Step-by-Step Explanation:
Step 1: Electrical conduction in liquids requires free ions.

Step 2: In distilled water:
Only very few ions are present due to self-ionisation:
\[ \mathrm{H_2O \rightleftharpoons H^+ + OH^-} \]
Ion concentration is extremely low \(\left(10^{-7}\,\mathrm{M}\right)\).
⇒ Hence, it does not conduct electricity.

Step 3: In rainwater:
Rainwater dissolves CO₂ and impurities from air.
\[ \mathrm{CO_2 + H_2O \rightarrow H_2CO_3} \]
Step 4: Carbonic acid ionises:
\[ \mathrm{H_2CO_3 \rightarrow H^+ + HCO_3^-} \]
Step 5: Presence of these ions increases conductivity.
⇒ Rainwater conducts electricity.

Final Conclusion:
Distilled water lacks sufficient ions, whereas rainwater contains dissolved ions, enabling electrical conduction.

Exam Significance:

Frequently asked conceptual question in CBSE board exams.
Tests understanding of role of ions in conductivity.
Important for competitive exams involving electrolytes and ionisation.
Builds foundation for topics like electrolysis and chemical conductivity.

Role of Water in Showing Acidic Behaviour

Conceptual Theory:

Acids show acidic behaviour only when they ionise in aqueous solution to produce hydrogen ions \(\mathrm{(H^+)}\). However, free \(\mathrm{H^+}\) ions do not exist independently; they combine with water molecules to form hydronium ions:

\[ \mathrm{HCl + H_2O \rightarrow H_3O^+ + Cl^-} \]

These \(\mathrm{H_3O^+}\) ions are responsible for acidic properties such as:

Turning blue litmus red
Conducting electricity
Reacting with metals and bases

In the absence of water, acids do not ionise, so no \(\mathrm{H^+}\) or \(\mathrm{H_3O^+}\) ions are formed.

Solution Roadmap:

Step 1: Recall that acidic behaviour depends on ionisation
Step 2: Identify role of water in ion formation
Step 3: Compare behaviour in presence and absence of water

Step-by-Step Explanation:
Step 1: Acids show acidic behaviour due to formation of \(\mathrm{H^+}\) (or \(\mathrm{H_3O^+}\)) ions.

Step 2: Ionisation of acids occurs only in aqueous solution:
\[ \mathrm{HCl + H_2O \rightarrow H_3O^+ + Cl^-} \]
Step 3: In the absence of water:
Acid molecules remain in molecular form and do not ionise.

Step 4: Since no ions are produced:

No change in litmus
No electrical conductivity
No acidic reactions

Step 5: Therefore, acidic behaviour is not observed.

Final Conclusion:
Acids do not show acidic behaviour without water because they cannot produce \(\mathrm{H^+}\) (or \(\mathrm{H_3O^+}\)) ions in the absence of water.

Exam Significance:

Very frequently asked conceptual question in CBSE exams.
Tests understanding of ionisation and role of water.
Important for competitive exams focusing on acid-base theory.
Helps avoid common misconception: presence of hydrogen ≠ acidity.

pH Scale, Nature of Solutions and Hydrogen Ion Concentration

Conceptual Theory:

The pH scale indicates the strength of acids and bases:

pH < 7 → Acidic
pH = 7 → Neutral
pH > 7 → Basic (alkaline)

Strength classification:

pH 0–3 → Strongly acidic
pH 4–6 → Weakly acidic
pH 8–10 → Weakly alkaline
pH 11–14 → Strongly alkaline

Hydrogen ion concentration is related to pH as:

\[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} \]

Hence, lower pH → higher \([\mathrm{H^+}]\).

Solution Roadmap:

Step 1: Note given pH values for A, B, C, D, E
Step 2: Classify each based on pH scale
Step 3: Match with required categories
Step 4: Arrange based on \([\mathrm{H^+}]\)

Step-by-Step Solution:
Step 1: Assign pH values:
A → 4
B → 1
C → 11
D → 7
E → 9

Step 2: Classify each:

B (pH 1) → Strongly acidic
A (pH 4) → Weakly acidic
D (pH 7) → Neutral
E (pH 9) → Weakly alkaline
C (pH 11) → Strongly alkaline

Step 3: Answer each part:

(a) Neutral → D
(b) Strongly alkaline → C
(c) Strongly acidic → B
(d) Weakly acidic → A
(e) Weakly alkaline → E

Step 4: Relation of pH and hydrogen ion concentration:
\[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} \]
Step 5: Arrange in increasing order of \([\mathrm{H^+}]\):
Higher pH → Lower \([\mathrm{H^+}]\)

Therefore order is:
\[ 11 < 9 < 7 < 4 < 1 \]
Corresponding solutions:
C < E < D < A < B

Exam Significance:

Very common CBSE board question combining pH scale + classification.
Tests conceptual clarity of logarithmic relation of pH.
Frequently appears in competitive exams as MCQs.
Strengthens understanding of acid-base strength and ion concentration.

Strength of Acids and Rate of Reaction with Metals

Conceptual Theory:

Acids react with metals to produce hydrogen gas. The rate of this reaction depends on the concentration of hydrogen ions \(\mathrm{(H^+)}\) present in the solution.

Strong acids (like HCl) completely ionise in water:

\[ \mathrm{HCl \rightarrow H^+ + Cl^-} \]

Weak acids (like acetic acid) partially ionise:

\[ \mathrm{CH_3COOH \rightleftharpoons H^+ + CH_3COO^-} \]

Therefore, strong acids produce more \(\mathrm{H^+}\) ions compared to weak acids of the same concentration.

Solution Roadmap:

Step 1: Identify type of acids (strong vs weak)
Step 2: Compare ionisation in both acids
Step 3: Relate \(\mathrm{H^+}\) concentration to reaction rate
Step 4: Conclude based on hydrogen gas evolution

Step-by-Step Solution:
Step 1: Reaction involved:
Magnesium reacts with acid to produce hydrogen gas:
\[ \mathrm{Mg + 2H^+ \rightarrow Mg^{2+} + H_2} \]
Step 2: Identify acids:
Test tube A → HCl (strong acid)
Test tube B → CH₃COOH (weak acid)

Step 3: Ionisation comparison:
HCl → completely ionises → more \(\mathrm{H^+}\)
CH₃COOH → partially ionises → fewer \(\mathrm{H^+}\)

Step 4: Effect on reaction:
More \(\mathrm{H^+}\) ions → faster reaction → more hydrogen gas bubbles (fizzing)

Step 5: Conclusion:
Test tube A will show more vigorous fizzing.

Final Answer:
Fizzing occurs more vigorously in test tube A (HCl) because it produces more \(\mathrm{H^+}\) ions than acetic acid.

Exam Significance:

Very important conceptual question on strong vs weak acids.
Frequently asked in CBSE board exams as reasoning-based questions.
Common in competitive exams testing reaction rates and ionisation.
Helps build understanding of chemical kinetics at basic level.

pH Change During Conversion of Milk into Curd

Conceptual Theory:

Fresh milk is slightly acidic with a pH around 6. When it is converted into curd, bacteria (Lactobacillus) ferment lactose (milk sugar) into lactic acid.

The formation of lactic acid increases the concentration of hydrogen ions \(\mathrm{(H^+)}\), which lowers the pH of the solution.

\[ \mathrm{Lactose \rightarrow Lactic\ Acid} \]

As acidity increases, the pH decreases (moves below 7).

Solution Roadmap:

Step 1: Identify initial pH of milk
Step 2: Understand bacterial fermentation
Step 3: Relate acid formation to increase in \(\mathrm{H^+}\)
Step 4: Conclude change in pH

Step-by-Step Explanation:
Step 1: Fresh milk has pH ≈ 6 (slightly acidic).

Step 2: During curd formation, bacteria convert lactose into lactic acid:
\[ \mathrm{Lactose \rightarrow Lactic\ Acid} \]
Step 3: Lactic acid increases hydrogen ion concentration:
\[ [\mathrm{H^+}] \uparrow \]
Step 4: Increase in \(\mathrm{H^+}\) leads to decrease in pH.

Step 5: Final pH becomes around 4.5 (more acidic).

Final Conclusion:
The pH decreases as milk turns into curd due to formation of lactic acid.

Exam Significance:

Frequently asked CBSE question linking chemistry with daily life.
Tests understanding of pH change and fermentation.
Important for competitive exams involving biochemical reactions.
Helps in understanding preservation and spoilage of food.

Effect of Baking Soda on Milk and Curd Formation

Conceptual Theory:

Fresh milk has a pH of about 6 (slightly acidic). When baking soda (NaHCO₃) is added, it reacts as a weak base and increases the pH of milk, making it slightly alkaline.

\[ \mathrm{NaHCO_3 \rightarrow Na^+ + HCO_3^-} \]

Lactic acid bacteria convert lactose into lactic acid, lowering the pH and causing milk to set into curd. This process depends strongly on suitable pH conditions.

Solution Roadmap:

Step 1: Identify effect of baking soda on pH
Step 2: Understand bacterial activity and fermentation
Step 3: Relate pH with curd formation
Step 4: Answer both parts logically

Step-by-Step Solution:

(a) Why is pH shifted to slightly alkaline?

Step 1: Fresh milk has pH ≈ 6 (slightly acidic).

Step 2: Baking soda is basic in nature.

Step 3: On adding NaHCO₃, it increases the pH of milk: \[ \mathrm{NaHCO_3 \rightarrow Na^+ + HCO_3^-} \]
Step 4: Increased pH makes milk slightly alkaline.

Step 5: This slows down bacterial activity, delaying spoilage.

Conclusion (a):
The milkman increases pH to prevent quick souring and extend freshness.

(b) Why does milk take longer to set as curd?

Step 1: Curd formation requires lactic acid production by bacteria.

Step 2: Lactic acid lowers pH: \[ \mathrm{Lactose \rightarrow Lactic\ Acid} \]
Step 3: Higher initial pH (due to baking soda) delays this decrease.

Step 4: Bacteria become less active in alkaline medium.

Step 5: Hence, acid production slows down and curd formation is delayed.

Conclusion (b):
Milk takes longer to set because increased pH slows bacterial fermentation and acid formation.

Exam Significance:

Very important CBSE reasoning-based question.
Links pH concept with real-life application.
Frequently asked in both board exams and competitive tests.
Helps understand food preservation and fermentation control.

Storage of Plaster of Paris (POP)

Conceptual Theory:

Plaster of Paris (POP) is calcium sulphate hemihydrate \(\mathrm{(CaSO_4 \cdot \frac{1}{2}H_2O)}\). It has a strong tendency to react with water (moisture) and convert into gypsum, which is a hard solid.

This reaction is called hydration and is irreversible under normal conditions.

Solution Roadmap:

Step 1: Identify chemical nature of POP
Step 2: Write hydration reaction
Step 3: Explain effect of moisture
Step 4: Conclude storage requirement

Step-by-Step Explanation:
Step 1: Plaster of Paris is chemically:
\[ \mathrm{CaSO_4 \cdot \frac{1}{2}H_2O} \]
Step 2: It reacts with water (moisture) to form gypsum:
\[ \mathrm{CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O \rightarrow CaSO_4 \cdot 2H_2O} \]
Step 3: Gypsum is a hard solid and cannot be reused as POP.

Step 4: If POP absorbs moisture from air:

It starts converting into gypsum
Loses its powdery nature
Becomes unusable for moulding

Step 5: Therefore, it must be protected from moisture.

Final Conclusion:
Plaster of Paris is stored in moisture-proof containers to prevent its reaction with water vapour, which would otherwise convert it into hard gypsum and make it unusable.

Exam Significance:

Frequently asked CBSE board question on POP and gypsum conversion.
Tests understanding of chemical reactions and storage conditions.
Important for competitive exams involving practical applications of chemistry.
Helps in understanding industrial uses of POP.

Neutralisation Reaction

Conceptual Theory:

A neutralisation reaction is a chemical reaction in which an acid reacts with a base to form salt and water. During this process, hydrogen ions \(\mathrm{(H^+)}\) from the acid combine with hydroxide ions \(\mathrm{(OH^-)}\) from the base to form water.

\[ \mathrm{H^+ + OH^- \rightarrow H_2O} \]

As a result, the acidic and basic properties cancel each other, leading to a neutral solution (pH ≈ 7).

Solution Roadmap:

Step 1: Define neutralisation reaction
Step 2: Write general form of reaction
Step 3: Provide balanced examples

Step-by-Step Answer:
Step 1: Definition:
A neutralisation reaction is a reaction in which an acid reacts with a base to form salt and water.

Step 2: General form:
\[ \mathrm{Acid + Base \rightarrow Salt + Water} \]
Step 3: Ionic explanation:
\[ \mathrm{H^+ + OH^- \rightarrow H_2O} \]
Step 4: Examples:

Example 1:
\[ \mathrm{HCl + NaOH \rightarrow NaCl + H_2O} \]
Example 2:
\[ \mathrm{H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O} \]

Exam Significance:

Very frequently asked definition-based question in CBSE exams.
Important for writing balanced chemical equations.
Common in competitive exams testing acid-base reactions.
Forms the foundation for understanding titration and pH control.

Uses of Washing Soda and Baking Soda

Conceptual Theory:

Washing soda and baking soda are important sodium compounds with wide industrial and domestic uses. Their properties arise from their basic nature and ability to react with acids or ions.

Solution Roadmap:

Step 1: Identify chemical names and formulas
Step 2: Recall their chemical properties
Step 3: State at least two important uses for each

Washing Soda (Sodium Carbonate, \(\mathrm{Na_2CO_3}\))

Step-by-Step Uses:

1. Water Softening:
Step 1: Hard water contains \(\mathrm{Ca^{2+}}\) and \(\mathrm{Mg^{2+}}\) ions.
Step 2: Washing soda reacts with these ions:
\[ \mathrm{Na_2CO_3 + Ca^{2+} \rightarrow CaCO_3 \downarrow} \]
Step 3: Insoluble calcium carbonate precipitates out, softening water.

2. Cleaning Agent:
Step 1: It is strongly basic in nature.
Step 2: It removes grease and oily stains effectively.
Step 3: Used in laundry, glass, and soap manufacturing industries.

Baking Soda (Sodium Hydrogen Carbonate, \(\mathrm{NaHCO_3}\))

Step-by-Step Uses:

1. Leavening Agent (Baking):
Step 1: On heating or reacting with acid, it decomposes:
\[ \mathrm{2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2} \]
Step 2: CO₂ gas makes cakes and bread soft and fluffy.

2. Antacid:
Step 1: It neutralises excess acid in the stomach:
\[ \mathrm{NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2} \]
Step 2: Provides relief from acidity and indigestion.

Exam Significance:

Very common CBSE board question (2–3 marks).
Tests memory + application of chemical compounds and their uses.
Frequently appears in competitive exams as direct MCQs.
Important for understanding real-life applications of chemistry.

Acids, Bases and Salts

Understanding pH and Litmus Behaviour

Acids Reacting with Carbonates

Neutralisation and Volume Relationships

Role of Antacids in Neutralising Stomach Acid

Acids Reacting with Metals (Salt + Hydrogen Formation)

Why Alcohols and Glucose Are Not Acids (Activity-Based Proof)

Electrical Conductivity of Water (Role of Ions)

Role of Water in Showing Acidic Behaviour

pH Scale, Nature of Solutions and Hydrogen Ion Concentration

Strength of Acids and Rate of Reaction with Metals

pH Change During Conversion of Milk into Curd

Effect of Baking Soda on Milk and Curd Formation

Storage of Plaster of Paris (POP)

Neutralisation Reaction

Uses of Washing Soda and Baking Soda

Chapter Complete!

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